如何虚拟继承解决“钻石” （多重继承）模糊？ [英] How does virtual inheritance solve the "diamond" (multiple inheritance) ambiguity?

问题描述

class A                     { public: void eat(){ cout<<"A";} }; 
class B: virtual public A   { public: void eat(){ cout<<"B";} }; 
class C: virtual public A   { public: void eat(){ cout<<"C";} }; 
class D: public         B,C { public: void eat(){ cout<<"D";} }; 

int main(){ 
    A *a = new D(); 
    a->eat(); 
} 

我理解钻石问题，上面的代码没有这个问题。

I understand the diamond problem, and above piece of code does not have that problem.

虚拟继承如何解决这个问题？

How exactly does virtual inheritance solve the problem?

我理解：
当我说 A * a = new D（）; 时，编译器想知道类型 D 可以分配给 A 类型的指针，但它有两个路径，它可以跟随，但不能自行决定。

What I understand: When I say A *a = new D();, the compiler wants to know if an object of type D can be assigned to a pointer of type A, but it has two paths that it can follow, but cannot decide by itself.

因此，虚拟继承如何解决问题（帮助编译器做出决定）？

So, how does virtual inheritance resolve the issue (help compiler take the decision)?

推荐答案

您需要：（可通过虚拟继承实现）

You want: (Achievable with virtual inheritance)

  D  
 / \  
B   C  
 \ /  
  A 

：（无虚拟继承会发生什么）

And not: (What happens without virtual inheritance)

  D  
 / \  
B   C  
|   |  
A   A 

虚拟继承意味着只有一个基础实例 A 类不是2.

Virtual inheritance means that there will be only 1 instance of the base A class not 2.

您的类型 D 指针（您可以在第一个图表中看到它们），一个用于 B ，一个用于 C c $ c> A 。 D 的对象大小增加，因为它现在存储2个指针;但是现在只有一个 A 。

Your type D would have 2 vtable pointers (you can see them in the first diagram), one for B and one for C who virtually inherit A. D's object size is increased because it stores 2 pointers now; however there is only one A now.

因此 B :: A 和 C :: A 是相同的，因此不能有 D 的模糊调用。如果你不使用虚拟继承，你有上面的第二个图。然后，任何对A成员的调用都会变得模糊，您需要指定要采用的路径。

So B::A and C::A are the same and so there can be no ambiguous calls from D. If you don't use virtual inheritance you have the second diagram above. And any call to a member of A then becomes ambiguous and you need to specify which path you want to take.

维基百科有另一个很好的范例和例子

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